<h2>The Forearm</h2>
Explanation:
The proximal end of the radius illustrates the relationship of form and function. The cup-like surface of the radial head articulates with the rounded shape of the capitulum. This forms a joint that allows for movement of elbows and forearms.
Radius and ulna are the two bones of the forearm. Their proximal ends articulate and fit into the cup like end of the humerus. This forms a synovial joint called the elbow joint. The movement of this joint allows the forearms to supinate and pronate.
Weathering (erosion), transportation, deposition, compaction. Essentially D is the correct answer.
The answers are as follows:
1. <span>An inhibitor has a structure that is so similar to the substrate that it can bond to the enzyme just like the substrate: t</span>his is called competitive inhibitor. A competitive inhibitor will compete with the substrate for the active site of the enzyme and bind to the active site, thus incapacitating the substrate from binding to the active site.
2. An inhibitor binds to a site on the enzyme that is not the active site: this is called non competitive inhibitors. Non competitive inhibitors bind to other site in the enzyme which is not the active site of the enzyme. The binding of the inhibitor changes the conformation of the enzyme as well as the active site, thus making it impossible for the substrate to bind to the enzyme effectively.
3. <span>usually, a(n) inhibitor forms a covalent bond with an amino acid side group within the active site, which prevents the substrate from entering the active site or prevents catalytic activity: this is called irreversible or permanent inhibition. Permanent inhibitors form covalent bonds with the enzyme and prevent substrate from binding to the enzyme.
4. T</span><span>he competitive inhibitor competes with the substrate for the ACTIVE SITE on the enzyme: The active site of an enzyme is the place where the substrate normally bind in order to activate a enzyme. Competitive inhibitors are those inhibitors that compete with the substrate for the active site of the enzyme and prevent the substrate from binding there.
5. W</span><span>hen the noncompetitive inhibitor is bonded to the enzyme, the shape of the ENZYME is distorted. The non competitive inhibitors are those inhibitors that bind to other places in the enzyme instead of the active site. The binding of the non competitive inhibitor usually distort the shape and the conformation of the enzyme thus preventing the substrate from binding to it effectively.
6. E</span><span>nzyme inhibitors disrupt normal interactions between an enzyme and its SUBSTRATE. The principal function of enzyme inhibitor is to prevent the substrate from binding to the appropriate enzyme. This is usually done in the human system in order to regulate the activities of enzymes.</span>
Answer:
B. At the edge of the plates
Explanation:
<u>Gravity </u>is the principal <u>driving force </u>of <u>plate tectonics </u>(second one is convection<u>)</u>. It causes different density plates to move on the Earth's surface. However, when a <u>denser plate coincides the less denser plate, the high density plate subducts</u> below the <u>lesser density plate</u>. The process, therefore, is called <u>subduction</u>. During this collision of plates, <u>shearing resistance increases</u> and all <u>pressures come at the edge of the plate</u>. The process continues and the lithosphere drags the rest of the plate. The portion of plate below the less denser plate then reaches the mantle. Here, the edge of plate is destroyed due to high temperature of mantle as well as pressure.
Answer:
Explanation:
To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.
The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.
So, en the exposed example:
- J and K are autosomal genes
- J and K are separated by 60 M.U.
- 60 M.U. means that there is 60% of recombination.
Cross) J K / j k x j k / j k
Gametes) JK Parental jk, jk, jk, jk
jk Parental
Jk Recombinant
jK Recombinant
One map unit equals 1% of recombination frequency. This means that every 100 meiotic products, one of them is a recombinant one.
1 M.U. -------------- 1% recombination
60 M.U. ------------ 60% recombination
30% Jk + 30% jK
100 M.U. - 60 M.U. = 40 M.U.
40M.U.--------------40 % Parental (Not recombinant)
20% JK + 20% jk
Punnet Square) JK jk Jk jK
jk JK/jk jk/jk Jk/jk jK/jk
J K / j k = 20%
j k / j k = 20%
J k / j k = 30%
j K / j k = 30%
